Calculate the enthalpy change when 100. g of ice at 0.0 °C is heated to liquid water at 50.0°C. (The heat of fusion for water is 333 J/g.)
1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ
1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ
Those are possible answers... Which is it?
q1 = heat to melt the ice.
q1 = mass of ice x heat of fusion.
q2 = heat to raise the temperature from 0 degrees C to 50 degrees C.
q2 = mass of water x specific heat water x (Tf - Ti)
Tf is final Temperature.
Ti is initial T.
Total heat = q1 + q2.
Post your work if you get stuck.
Never mind, I got it:
100 grams of ice times 333 J/g = 33.3 kJ of heat to melt all the ice.
Next, raising temperature of liquid, so 100 g *4.184 J/g *50= 20.92 kJ
Thus A is best answer
right. In fact, I obtained 54.2 kJ.
Well, let's start by getting rid of the options that are just plain wrong. I mean, who in their right mind would think the answer is 54,220 kJ? That's like the enthalpy change of the entire Arctic melting!
And 20.9 kJ sounds way too low for all that ice to melt, don't you think? I mean, it's a whole 100 grams we're talking about!
So now we have two options left. 54.2 kJ and 33.3 kJ. Hmmm... this is a tough one. You know, I could really go for a nice glass of water right now. But since I can't actually drink water, I'll just have to give you the answer.
The correct answer is... drumroll, please... 54.2 kJ! Ding, ding, ding! We have a winner! Congrats!
So yeah, the enthalpy change when 100 g of ice at 0.0 °C is heated to liquid water at 50.0°C is 54.2 kJ. Enjoy your melted ice!
To calculate the enthalpy change, we need to consider the different steps involved in the process: heating the ice from 0.0 °C to its melting point, melting the ice, and then heating the resulting water from 0.0 °C to 50.0 °C.
Step 1: Heating the ice from 0.0 °C to its melting point.
To calculate the heat absorbed in this step, we can use the formula:
Q = m * C * ΔT
Where:
Q is the heat absorbed/released
m is the mass of the substance
C is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, the specific heat capacity of ice is approximately 2.09 J/g·°C (specific heat capacity of water).
The temperature change (ΔT) is the final temperature minus the initial temperature:
ΔT = (0.0 °C - (-0.0 °C)) = 0.0 °C
Using the formula, we find:
Q1 = 100. g * 2.09 J/g·°C * 0.0 °C = 0 J
Step 2: Melting the ice.
To calculate the heat absorbed in this step, we use the heat of fusion for water, which is given as 333 J/g. The mass of the ice being melted is 100 g, so the total heat absorbed is:
Q2 = 100. g * 333 J/g = 33,300 J
Step 3: Heating the resulting water from 0.0 °C to 50.0 °C.
Using the specific heat capacity of water (4.18 J/g·°C), we calculate:
Q3 = 100. g * 4.18 J/g·°C * (50.0 °C - 0.0 °C) = 209,000 J
To find the total enthalpy change, we sum up the heats absorbed in each step:
Total enthalpy change = Q1 + Q2 + Q3 = 0 J + 33,300 J + 209,000 J = 242,300 J = 242.3 kJ
Therefore, the correct answer is 242.3 kJ, which is not among the given options.